Algebra 1 Question on Rate, Distance, and Time

Question

It is 74.4 miles from Dwane and Patti's house to Martha's house. Patti leaves home at 10:00 and drives toward Martha's house. Dwane and Martha set off half an hour later from their own houses and drive toward the other's house. The average speed of the three friends is 54 mph and they all meet at 11:06. How fast does each drive?

Answer 1

*—————————————————*
D&P                   74.4 miles                    M

10:00 P starts
10:30 D & M start
11:06 all meet (at same point)

dD = dP is distance D & P traveled
dM + dD = 74.4 miles
(rD + rP + rM)/3 = 54 mph

11:06 - 10:00 = 1.1 hours = tP
11:06 - 10:30 = 0.6 hours = tD = tM

dD = rD*tD = 0.6 rD
dM = rM*tM = 0.6 rM
dP = rP*tP = 1.1 rP

0.6 rD = 1.1 rP; from dD = dP
0.6 rM + 0.6 rD = 74.4; from dM + dD = 74.4
(rD + rP + rM)/3 = 54; given

Eqn-1:    rD +     rP +   rM = 162
Eqn-2: 6 rD - 11 rP + 0 rM = 0
Eqn-3:    rD +  0 rP +   rM = 124

Eqn-1 - Eqn-3:

rP = 162 - 124 = 38 mph

Substitute in Eqn-2:

6 rD = 11(38)

rD = 11(2*19)/(2*3) = 11(19)/3 = 209/3 = 69 2/3 mph

Substitute in Equ-1:

209/3 + 38 + rM = 162

rM = 124 - (69 + 2/3) = 55 - 2/3 = 54 1/3 mph

Summary:

rD = 69 2/3 mph
rP = 38 mph
rM = 54 1/3 mph

check:

rD * tD = (69+2/3 mph)(0.6 hours) = 41.4 + 0.4 = 41.8 miles

rP * tP = (38 mph)(1.1 hours) = 41.8 miles = rD * tD √

rM * tM = (54 1/3 mph)(0.6 hours) = 32.4 + 0.2 = 32.6 miles

41.8 + 32.6 = 74.4 miles √

Answer 2

Sorry for my error. Please see Steve's answer.

 

Thanks for your correction Steve.  I assumed they traveled the same distance which was the flaw in my set up. Thanks again.